Question: The value of $\sqrt{96}$ lies between which two consecutive integers ? Integers that appear in order when counting, for example 2 and 3.
Solution: Consider the perfect squares near $96$ . [ What are perfect squares? Perfect squares are integers which can be obtained by squaring an integer. The first 13 perfect squares are: $ 1,4,9,16,25,36,49,64,81,100,121,144,169$ $81$ is the nearest perfect square less than $96$ $100$ is the nearest perfect square more than $96$ So, we know $81 < 96 < 100$ So, $\sqrt{81} < \sqrt{96} < \sqrt{100}$ So $\sqrt{96}$ is between $9$ and $10$.